A current of 0.25 A flows in the main circuit. Now the resistance R is disconnected and then connected across the 4 ohm resistance. The current in the curcuit now is

 From above circuit diagram R_net= 2R/(2+R)+4+3,      = (9R+4)/(2+R),
ACCORDING TO CONDITION I=0.25, V= 2VOLT
                      0.25= 2(2+R)/(9R+4)       (​Since I=V/R)
            on solving for R,
            R = 2 ohm
now as per question this resistor is connected across resistor 4 ohm...
net R=  2+3+4*2/(4+2)= 6.333ohm
      V_given=2volt
so I = V/R  =  2/6.33 
      I=0.31A